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Lecture Outline Chapter 10 Motion in a Plane © 2015 Pearson Education, Inc. Slide 10-1 Chapter 10: Motion in a plane Chapter Goal: Develop the tools that allow us to deal with motion that takes place in two dimensions. © 2015 Pearson Education, Inc. Slide 10-2 Section 10.1: Straight is a relative term • The ball is dropped from a pole attached to a cart that is moving to the right at a constant speed. https://www.youtube.com/watch?v=mYH_nODWkqk Skateboarding Frame of Reference Demonstration © 2015 Pearson Education, Inc. Slide 10-3 Section 10.1: Straight is a relative term • (a) The ball falls to the ground in a straight line if observed from the cart’s reference frame. • (b) The ball has a horizontal displacement in addition to the straight downward motion when observed from Earth’s reference frame. The motion of the ball in Earth’s reference frame can be broken down into two parts: • Free fall in the vertical direction • Constant velocity motion in the horizontal direction © 2015 Pearson Education, Inc. g g Slide 10-4 Section 10.1 Clicker Question 1 A passenger in a speeding train drops a peanut. Which is greater? 1. The magnitude of the acceleration of the peanut as measured by the passenger. 2. The magnitude of the acceleration of the peanut as measured by a person standing next to the track. 3. Neither, the accelerations are the same to both observers. Slide 10-5 Section 10.1 Clicker Question 1 A passenger in a speeding train drops a peanut. Which is greater? 1. The magnitude of the acceleration of the peanut as measured by the passenger. 2. The magnitude of the acceleration of the peanut as measured by a person standing next to the track. 3. Neither, the accelerations are the same to both observers. Slide 10-6 Section 10.2: Vectors in a plane • To analyze the motion of an object moving in a plane, we need to define two reference axes, x and y. • The ball’s displacement in Earth’s reference frame is the vector sum of the horizontal displacement Dx and the vertical displacement Dy. © 2015 Pearson Education, Inc. Slide 10-7 Section 10.2: Vectors in a plane • The vector sum of two vectors in a plane is obtained by placing the tail of the second vector at the head of the first vector. • To subtract a vector b from a vector a, reverse the direction b of and then add the reversed b to a. © 2015 Pearson Education, Inc. Slide 10-8 Section 10.2: Vectors in a plane • Any vector A can be decomposed to component vectors Ax and Ay along the axes of some conveniently chosen set of mutually perpendicular axes, called a rectangular coordinate system. • The procedure for decomposing a vector is shown below. © 2015 Pearson Education, Inc. Slide 10-9 Section 10.2: Vectors in a plane • The figure below shows the decomposition of a vector A in three coordinate systems. • You need to choose the coordinate system that best suits the problem at hand. © 2015 Pearson Education, Inc. Slide 10-10 Section 10.2: Vectors in a plane • The displacement, instantaneous velocity, and acceleration of the ball in the previous section is shown below. • Notice that the instantaneous velocity and acceleration are not in the same direction. What does this mean? © 2015 Pearson Education, Inc. Slide 10-11 Section 10.2: Vectors in a plane • Decompose the acceleration vector into two components: one parallel to the instantaneous velocity, and one perpendicular to it. v^ v|| v|| F = mg • The component of the acceleration of the ball parallel to the instantaneous velocity changes the speed; the component of acceleration perpendicular to the instantaneous velocity changes the direction of the velocity but not its magnitude. • Velocity component parallel to acceleration changes. • Velocity component perpendicular to acceleration stays unchanged. © 2015 Pearson Education, Inc. Slide 10-12 Checkpoint 10.2 10.2 In Figure 10.10, the ball’s instantaneous velocity u does not point in the same direction as the displacement Dr (it points above the final position of the ball). Why? © 2015 Pearson Education, Inc. Slide 10-13 Section 10.3: Decomposition of forces • The figure shows a brick lying on a horizontal plank and then the plank is gently tilted. At first, due to friction, it remains at rest. • When the angle of incline exceeds a θmax the brick accelerates down the incline. • Then, the vector sum of the forces exerted on the brick must also point down the incline. © 2015 Pearson Education, Inc. Slide 10-14 Section 10.3: Decomposition of forces • Because the brick is constrained to move along the surface of the plank, it make sense to choose the x axis along surface, as illustrated in the figure. • The force components parallel to the surface are called tangential components. • The force components perpendicular to the surface are called normal components. © 2015 Pearson Education, Inc. Slide 10-15 Section 10.3: Decomposition of forces • The brick problem suggests which axes we should choose in a given problem: • If possible, choose a coordinate system such that one of the axes lies along the direction of the acceleration of the object under consideration. © 2015 Pearson Education, Inc. Slide 10-16 Lecture Outline Chapter 10 Motion in a Plane © 2015 Pearson Education, Inc. Slide 10-17 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing Using a rope, you pull a friend sitting on a swing (Figure 10.16). (a) As you increase the angle θ, does the magnitude of the force Frpc required to hold your friend in place increase or decrease? Frpc © 2015 Pearson Education, Inc. Slide 10-18 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing (cont.) (b) Is the magnitude of Frpc larger than, equal to, or smaller than the magnitude of the gravitational force FEpG exerted by Earth on your friend? (Consider the situation for both small and large values of θ.) Claim: There is an angle at which © 2015 Pearson Education, Inc. Frpc FEpG Slide 10-19 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing (cont.) (c) Is the magnitude of the force Fspc exerted by the swing on your friend larger than, equal to, or smaller than FEpG ? Fspc © 2015 Pearson Education, Inc. FEpG Slide 10-20 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing ❶ GETTING STARTED Begin by drawing a free-body diagram of your friend (Figure 10.17). Three forces are exerted on him: FEpG , the force of gravity directed vertically downward, the horizontal force Frpc exerted by the rope, and a force Fspc exerted by the swing seat. © 2015 Pearson Education, Inc. Slide 10-21 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing ❶ This latter force Fspc is exerted by the suspension point via the chains of the swing and is thus directed along the chains. Choose a horizontal x axis and a vertical y axis, so that two of the three forces lie along axes. Because your friend’s acceleration is zero, the vector θ sum of the forces must be zero. © 2015 Pearson Education, Inc. Slide 10-22 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing ❸ EXECUTE PLAN (a) Decompose Fspc into x and y components. The forces must add to zero along both axes. Therefore, Fspc y must be equal in magnitude to FEpG , the c downward force of gravity. Likewise, Fsp x must be equal in magnitude to Frpc , the horizontal force the rope exerts on your friend. θ © 2015 Pearson Education, Inc. Slide 10-23 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing ❸ EXECUTE PLAN Next, draw a second free-body diagram for a larger angle θ. As θ increases, Fspc y must remain FEpG to equal in magnitude (otherwise your friend would accelerate vertically). Increasing θ while keeping constant Fspc y requires the magnitude of to increase. © 2015 Pearson Education, Inc. Slide 10-24 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing (a) As you increase the angle θ, does the magnitude of the force Frpc required to hold your friend in place increase or decrease? ❸ EXECUTE PLAN Because your friend is at rest, the forces in the horizontal direction must add up to zero and so Frpc = Fspc x . So if the magnitude of Fspc x increases, the magnitude of Frpc must increase, too. ✔ © 2015 Pearson Education, Inc. Slide 10-25 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing (b) Is the magnitude of Frpc larger than, equal to, or smaller than the magnitude of the gravitational force FEpG ,exerted by Earth on your friend? ❸ EXECUTE PLAN (b) From Figure 10.18, tanq = Fspc x / Fspc y . For θ < 45, tan θ < 1, and so Fspc x < Fspc y . Because Fspc y = FEpG and F c = F c , I discover that for sp x rp θ < 45, When θ > 45, tan θ > 1, and F c > F c sp x sp y so θ > 45, Slide 10-26 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing (c) Is the magnitude of the force Fspc exerted by the swing on your friend larger than, equal to, or smaller than FEpG ? ( ) ( ). 2 + F ❸ EXECUTE PLAN (c) F = F and F = F Therefore, Fspc must always be larger than F when θ ≠ 0. ✔ c sp y © 2015 Pearson Education, Inc. G Ep c sp G Ep c sp x c sp y 2 Slide 10-27 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing ❹ EVALUATE RESULT I know from experience that you have to pull harder to move a swing farther from its equilibrium position, and so my answer to part a makes sense. With regard to part b, when the swing is at rest at 45, the forces Frpc and FEpG on your friend make the same angle with the force Fspc , and so Frpc and FEpG should be equal in magnitude. Slide 10-28 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing ❹ EVALUATE RESULT The force of gravity is independent of the angle, but the force exerted by the rope increases with increasing angle, and so it makes sense that for angles larger than 45, In part c, because the vertical component of the force Fspc exerted by the seat on your friend always has to be equal to the force of gravity, adding a horizontal component makes Fspc larger than . θ > 45 Fspc Frpc q Slide 10-29 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing tan q = Fspc x / Fspc y q Frpc Fspc Fspc q Slide 10-30 Section 10.3: Decomposition of forces Example 10.2 Pulling a friend on a swing tan q = Fspc x / Fspc y Iron cross q Frpc Fspc Fspc q Slide 10-31 Section 10.6: Vector algebra • As illustrated in the figure, the decomposition of an arbitrary vector A can be written as: A = Ax + Ay = Ax iˆ + Ay ĵ where iˆ and ĵ represent unit vectors along x and y axes. • Using the Pythagorean theorem we can find the magnitude of A as A º A = + Ax2 + Ay2 • The angle of A with respect to the Ay x axis is tan q = Ax © 2015 Pearson Education, Inc. Slide 10-32 Section 10.6: Vector algebra • As illustrated in the figure, the decomposition of vectors is particularly useful when adding or subtracting them: R = ( Ax + Bx )iˆ + ( Ay = By ) ĵ Rx = Ax + Bx Ry = Ay + By © 2015 Pearson Education, Inc. Slide 10-33 Section 10.7: Projectile motion in two dimensions • The position vector of an object moving in two dimensions is r = xiˆ + yĵ • The object’s instantaneous velocity is Dx ˆ Dy u = lim i + lim ĵ = u xiˆ + u y ĵ Dt®0 Dt Dt®0 Dt so that, dx dy ux = and u y = dt dt • Similarly, instantaneous acceleration components are du y d 2 y du x d 2 x ax = = 2 and a y = = 2 dt dt dt dt • Decomposing vectors into components allows us to separate motion in a plane into two one-dimensional problems. © 2015 Pearson Education, Inc. Slide 10-34 Section 10.7: Projectile motion in two dimensions © 2015 Pearson Education, Inc. Slide 10-35 Section 10.7: Projectile motion in two dimensions • Consider the motion of a ball launched straight up from a cart moving at a constant velocity (see the figure on the previous slide). • The resulting two-dimensional projectile motion in Earth’s reference frame is the curved trajectory shown in the figure. • The balls has a constant downward acceleration of ay = –g. u x,i q u y,i ay = –g ui = u x,iiˆ + u y,i ĵ u y,i tan q = u x,i q © 2015 Pearson Education, Inc. Slide 10-36 Section 10.7: Projectile motion in two dimensions • In Earth’s reference frame, the ball’s initial velocity is u = u iˆ + u ĵ i x,i y,i • The ball’s launch angle relative to the x axis is u y,i tan q = u x,i • Using ax = 0 and ay = –g in Equations 3.4 and 3.8, υx,f = υx,i (constant velocity) υy,f = υy,i – gΔt (constant acceleration) xf = xi + υx,i Δt (constant velocity) u x,i q u y,i yf = yi + u y,i Dt - 12 g(Dt)2 (constant acceleration) ay = –g y q x © 2015 Pearson Education, Inc. Slide 10-37 Section 10.7: Projectile motion in dimensions Example 10.5 Position of highest point The ball is launched from the origin of an xy coordinate system. Write expressions giving the ball’s rectangular coordinates at the top of its trajectory, in terms of its initial speed υi and the acceleration due to gravity g. xf = xi + υx,i Δt (constant velocity) u x,i yf = yi + u y,i Dt - 12 g(Dt)2 (constant acceleration) q u y,i Need to find time to get to the top: Δttop ay = –g y q © 2015 Pearson Education, Inc. x Slide 10-38 Section 10.7: Projectile motion in two dimensions Example 10.5 Position of highest point (cont.) ❶ Because the ball is launched from the origin, xi = 0 and yi = 0. As the ball passes through its highest position, υy reverses sign from positive to negative, so at the highest point υy,f = 0. υy,f = υy,i – gΔttop = 0 Solving for Δttop then yields Dttop = æu ö u u xtop = 0 + u x,i çç y,i ÷÷ = x,i y,i g è g ø u y,i g æu ö æu ö u u u y,i y,i 1 1 1 ÷÷ - 2 g çç ÷÷ = ytop = 0 + u y,i çç -2 =2 g g g è g ø è g ø 2 © 2015 Pearson Education, Inc. 2 y,i 2 y,i . u x,i q u y,i ay = –g y 2 y,i q x Slide 10-39 Section 10.7: Projectile motion in two dimensions Example 10.5 Position of highest point (cont.) ❹ EVALUATE RESULT Because the ball moves at constant velocity in the horizontal direction, the x coordinate of the highest position is the horizontal velocity component υx,i multiplied by the time interval it takes to æu ö u u reach the top. ÷÷ = xtop = 0 + u x,i çç g è ø y,i x,i y,i © 2015 Pearson Education, Inc. u y,i ay = –g g The y coordinate of the highest position does not depend on υx,i because the vertical and horizontal motions are independent of each other. 2 ytop = u x,i q 1 2 u y,i y q x g Slide 10-40 Section 10.7: Projectile motion in two dimensions Example 10.6 Range of projectile How far from the launch position is the position at which the ball is once again back in the cart? (This distance is called the horizontal range of the projectile.) yf = yi + u y,it - 12 gt 2 (constant acceleration) 0 = 0 + u y,it - 12 g t 2 = t(u y,i - 12 g t) t = 0 or t = 2v y,i g æ 2u y,i ö 2u x,iu y,i xf = 0 + u x,i ç = . ÷ g è g ø u x,i q u y,i ay = –g y q x © 2015 Pearson Education, Inc. Slide 10-41 Section 10.7: Projectile motion in two dimensions Example 10.6 Range of projectile (cont.) ❸ Δttop = υy,i/g. The time interval spent in the air is Δtflight = 2υy,i/g. æ 2u y,i ö 2u x,iu y,i xf = 0 + u x,i ç = . ÷ g è g ø u x,i = ui cos(q ) xf = 2u x,iu y,i g u x,i q range u y,i u y,i = ui sin(q ) 2u i2 sin(q )cos(q ) u i2 sin(2q ) = = g g xf ay = –g range q 0 0 45 q 90 xf without air friction Slide 10-42 Section 10.7: Projectile motion in two dimensions Example 10.6 Range of projectile (cont.) y = 9o E = 2.6 cm optimal angle: 25o Slide 10-43 Section 10.7: Projectile motion in two dimensions Example 10.6 Range of projectile (cont.) https://www.youtube.com/watch?v=y9CEuJ5e2cM Garciaparra 30sec – 1:30 https://www.youtube.com/watch?v=6VMVQbIq0V8 Ortiz optimal angle: 25o Manny Ramirez Slide 10-44 Lecture Outline Chapter 10 Motion in a Plane © 2015 Pearson Education, Inc. Slide 10-45 Section 10.8: Collisions and momentum in two dimensions • Momentum conservation states that the momentum of an isolated system of colliding objects does not change, or Dp = 0. • Momentum is a vector, so in two dimensions momentum change must be expressed in terms of the components. • Conservation of momentum in two dimensions is given by Δpx = Δp1x + Δp2x = m1(υ1x,f – υ1x,i) + m2(υ2x,f – υ2x,i) = 0 Δpy = Δp1y + Δp2y = m1(υ1y,f – υ1y,i) + m2(υ2y,f – υ2y,i) = 0. u1,f u1,i u 2,f u 2,i © 2015 Pearson Education, Inc. Slide 10-46 Section 10.4: Friction • Consider the contact force exerted by the floor on the cabinet: • The normal component is called the normal force and the tangential component is the force of friction. • To understand the difference between normal and frictional forces, consider the brick on a horizontal wooden plank. • As seen in the figure, the normal force takes on whatever value to balance the net downward force, up to the breaking point. © 2015 Pearson Education, Inc. Slide 10-47 Section 10.4: Friction • Now consider gently pushing the brick to the right, as shown. • The horizontal frictional force is caused by microscopic bonds between the surfaces in contact. • As you push the brick, the net effect of these microscopic forces is to hold the brick in place. • As you increase your push force, this tangential component of the contact force grows. © 2015 Pearson Education, Inc. Slide 10-48 Section 10.4: Friction • The friction exerted by the surfaces that are not moving relative to each other is called static friction. • When the horizontal push force exceeds the maximum force of static friction, the brick will accelerate. • The friction force exerted by the surfaces when they move relative to each other is call kinetic friction, which is caused by transient microscopic bonds between the two surfaces. © 2015 Pearson Education, Inc. Slide 10-49 Section 10.4 Clicker Question 3 You push horizontally on a crate at rest on the floor, gently at first and then with increasing force until you cannot push harder. The crate does not move. What happens to the force of static friction between crate and floor during this process? 1. 2. 3. 4. 5. It is always zero. It remains constant in magnitude. It increases in magnitude. It decreases in magnitude. More information is needed to decide. © 2015 Pearson Education, Inc. Slide 10-50 Section 10.4 Clicker Question 3 You push horizontally on a crate at rest on the floor, gently at first and then with increasing force until you cannot push harder. The crate does not move. What happens to the force of static friction between crate and floor during this process? 1. 2. 3. 4. 5. It is always zero. It remains constant in magnitude. It increases in magnitude. It decreases in magnitude. More information is needed to decide. © 2015 Pearson Education, Inc. Slide 10-51 Section 10.5: Work and Friction • Like normal force, the force of static friction is an elastic force that causes no irreversible change. • The force of static friction causes no energy dissipation. • Kinetic friction does cause irreversible change, including causing microscopic damage to the surfaces. • The force of kinetic friction is not an elastic force and so causes energy dissipation. Slide 10-52 Section 10.5: Work and Friction • Two examples where the object is accelerated by static friction. © 2015 Pearson Education, Inc. Slide 10-53 Section 10.9: Work as the product of two vectors • Consider a block sliding down an incline as shown (ignore friction). • The x component of the force of gravity is what causes the block to accelerate G FEbx = FEbG sinq = +mg sinq • The magnitude of the resulting acceleration is å Fx åGEb x +mg sinq ax = = = = +g sinq m m m © 2015 Pearson Education, Inc. Slide 10-54 Section 10.9: Work as the product of two vectors • As illustrated in the figure, the block begins at rest (υx,i = 0) at position xi = 0 and drops a height h. • The displacement along the incline is Δx = +h/sinθ. • The time the block takes to cover this distance is obtained from Equation 3.7: Dx = u x,i Dt + 12 ax (Dt)2 = 0 + 12 g sinq (Dt)2 2h from which we obtain (Dt) = g sin 2 q 2 © 2015 Pearson Education, Inc. Slide 10-55 Section 10.9: Work as the product of two vectors Using Equation 3.4 and Equation 10.25, the blocks final kinetic energy is then 1 2 2 mu x,f = 12 m(u x,i + ax Dt 2 ) = 12 m(0 + g sinq Dt)2 = 12 mg 2 sin 2 q (Dt)2 Substituting Equation 10.28 into Equation 10.29, we obtain 2h ö 2 2 2 æ 1 1 mu x,f = 2 mg sin q ç = mgh 2 2 ÷ è g sin q ø Because the blocks initial kinetic energy is zero, we get 2 W = DK = K f – K i = 12 mu x,f – 0 = mgh We can get the same result by using the work equation: æ h ö W = FEbG x Dx = (+mg sinq ) ç + = mgh ÷ è sin q ø Slide 10-56 Section 10.9: Work as the product of two vectors • If the angle between A and B is ϕ, the scalar product of the two vectors is A× B º ABcosf • The scalar product is commutative: A× B º B × A = ABcosf = BAcosf © 2015 Pearson Education, Inc. Slide 10-57 Section 10.9 Clicker Question 8 Under what circumstances could the scalar product of two vectors be zero? 1. When the vectors are parallel to each other 2. When the vectors are perpendicular to each other 3. When the vectors make an acute angle with each other 4. None of the above © 2015 Pearson Education, Inc. Slide 10-58 Section 10.9 Clicker Question 8 Under what circumstances could the scalar product of two vectors be zero? 1. When the vectors are parallel to each other 2. When the vectors are perpendicular to each other 3. When the vectors make an acute angle with each other 4. None of the above © 2015 Pearson Education, Inc. Slide 10-59 Section 10.9: Work as the product of two vectors • We can write work as a scalar product: W = F × DrF (constant nondissipative force) • The normal force does no work, because ϕ = 90 between the normal force and the force displacement. • The work done by gravity on the block is f æ ö h h W = FEbG × DrF = (mg) ç cos f è sinq ÷ø where θ is the angle of the incline. • However, since ϕ = π/2 – θ, we have W = mgh. • If force and force displacement are opposite (ϕ = 180), then work is negative. © 2015 Pearson Education, Inc. Slide 10-60 Section 10.9: Work as the product of two vectors • Let us generalize the previous result to forces that are not constant, F(r ). • Start by subdividing the force displacement DrF into many small fragments d rFn . Work done by a variable force F(r ) over a small displacement is Wn » F(rn ) × d rFn • The work done over the entire force displacement is: rf W = ò F(r ) × dr (variable nondissipative force) ri © 2015 Pearson Education, Inc. Slide 10-61 Section 10.10: Coefficients of Friction • Let us now develop a quantitative description of friction: • The figure illustrates the following observations. • The maximum force of static friction exerted by a surface on an object is proportional to the force with which the object presses the surface. • The force of static friction does not depend on the contact area. © 2015 Pearson Education, Inc. Slide 10-62 Section 10.10: Coefficients of Friction • As we have just determined, the maximum magnitude of the static friction force must be proportional to Fsbn , as seen in the figure. • Therefore, for any two surfaces 1 and 2 we have (F12s )max = ms F12n • The unitless proportionality constant μs is called the coefficient of static friction. • This upper limit means that the magnitude of static friction must obey the following condition: F12s £ ms F12n © 2015 Pearson Education, Inc. Slide 10-63 Section 10.10: Coefficients of Friction © 2015 Pearson Education, Inc. Slide 10-64 Section 10.10 Clicker Question 9 In a panic situation, many drivers make the mistake of locking their brakes and skidding to a stop rather than applying the brakes gently. A skidding car often takes longer to stop. Why? 1. This is a misconception; it doesn’t matter how the brakes are applied. 2. The coefficient of static friction is larger than the coefficient of kinetic friction. 3. The coefficient of kinetic friction is larger than the coefficient of static friction. © 2015 Pearson Education, Inc. Slide 10-65 Section 10.10 Clicker Question 9 In a panic situation, many drivers make the mistake of locking their brakes and skidding to a stop rather than applying the brakes gently. A skidding car often takes longer to stop. Why? 1. This is a misconception; it doesn’t matter how the brakes are applied. 2. The coefficient of static friction is larger than the coefficient of kinetic friction. 3. The coefficient of kinetic friction is larger than the coefficient of static friction. © 2015 Pearson Education, Inc. Slide 10-66 Section 10.10: Coefficients of Friction Example 10.8 Pulling a friend up a hill A hiker is helping a friend up a hill that makes an angle of 30 with level ground. The hiker, who is farther up the hill, is pulling on a cable attached to his friend. The cable is parallel to the hill so that it also makes an angle of 30 with the horizontal. If the coefficient of static friction between the soles of the hiker’s boots and the surface of the hill is 0.80 and his inertia is 65 kg, what is the maximum magnitude of the force he can exert on the cable without slipping? © 2015 Pearson Education, Inc. Slide 10-67 Section 10.10: Coefficients of Friction Example 10.8 Pulling a friend up a hill (cont.) ❶ GETTING STARTED The forces exerted on the hiker are a force Fchc exerted by the cable and directed down the hill (this force forms an interaction pair with the force the hiker exerts on the cable), the force of gravity FEhG , and the contact force exerted by the surface of the hill on the hiker. © 2015 Pearson Education, Inc. Slide 10-68 Section 10.10: Coefficients of Friction Example 10.8 Pulling a friend up a hill (cont.) ❶ GETTING STARTED This last force has two components: the normal force Fshn and the force of static friction Fshs. The force of static friction is directed up the hill. If the hiker is not to slip, his acceleration must be zero, and so the forces in the normal (y) and tangential (x) directions must add up to zero. © 2015 Pearson Education, Inc. Slide 10-69 Section 10.10: Coefficients of Friction Example 10.8 Pulling a friend up a hill (cont.) ❸ Because the magnitude of the force of gravity exerted on the hiker is mg, I have in the y direction G n n F = F + F = –mg cos q + F = ma y = 0 å y Eh y sh sh Fshn = mg cosq where the minus sign in –mg cos θ indicates that FEhG y is in the negative y direction. © 2015 Pearson Education, Inc. Slide 10-70 Section 10.10: Coefficients of Friction Example 10.8 Pulling a friend up a hill (cont.) ❸ Likewise, I have in the x direction G s c F = F + F – F å x Eh x sh ch = –mg sinq + Fshs – Fchc = max = 0 Fshs = mg sinq + Fchc © 2015 Pearson Education, Inc. Slide 10-71 Section 10.10: Coefficients of Friction Example 10.8 Pulling a friend up a hill (cont.) ❸ EXECUTE PLAN If the hiker is not to slip, the force of static friction must not exceed its maximum value. Substituting Eqs. 1 and 2 into Inequality 10.54, I get mg sin q + Fchc £ ms (mg cosq ) Fchc £ mg( ms cosq – sin q ). © 2015 Pearson Education, Inc. Slide 10-72 Section 10.10: Coefficients of Friction Example 10.8 Pulling a friend up a hill (cont.) ❸ EXECUTE PLAN The problem asks me about the magnitude Fhcc of the force exerted by the hiker on the cable, which is the same as the magnitude Fchc of the force exerted by the cable on the hiker, so Fhcc £ mg( ms cosq – sinq ). © 2015 Pearson Education, Inc. Slide 10-73 Section 10.10: Coefficients of Friction Example 10.8 Pulling a friend up a hill (cont.) ❸ Substituting the values given, Fhcc £ (65 kg)(9.8 m/s 2 )(0.80 cos 30° – sin30°) =1.2 ´ 102 N. © 2015 Pearson Education, Inc. Slide 10-74 Section 10.10: Coefficients of Friction Example 10.8 Pulling a friend up a hill (cont.) ❹ EVALUATE RESULT A force of 1.2 × 102 N corresponds to the gravitational force exerted by Earth on an object that has an inertia of only 12 kg, and so the hiker cannot pull very hard before slipping. I know from experience, however, that unless I can lock a foot behind some solid object, it is impossible to pull hard on an inclined surface, and thus the answer I obtained makes sense. © 2015 Pearson Education, Inc. Slide 10-75 Lecture Outline Chapter 10 Motion in a Plane © 2015 Pearson Education, Inc. Slide 10-76 Section 10.10: Coefficients of Friction • Once surfaces start to slip relative to each other, the force of kinetic friction is relative to the normal force F12k = mk F12n where μk is called the coefficient of kinetic friction. • The kinetic coefficient μk is always smaller then μs. © 2015 Pearson Education, Inc. Slide 10-77 Chapter 10: Summary Concepts: Vectors in two dimensions • To add two vectors A and B, place the tail of B at the head of A.Their sum is the vector drawn from tail of A to the head of B. • To subtract B from A, reverse the direction of B and then add the reversed B to A. • Any vector A can be written as A = Ax + Ay = Ax iˆ + Ay ĵ where Ax and Ay are the components of A along the x and y axes. © 2015 Pearson Education, Inc. Slide 10-78 Chapter 10: Summary Quantitative Tools: Vectors in two dimensions • The magnitude of any vector A is A º A = + Ax2 + Ay2 and the angle θ that A makes with the positive x axis is given by Ay tan q = . Ax • If R = A + B is the vector sum of A and B, the component of R are Rx = Ax + Bx Ry = Ay + By . • The scalar product of two vectors A and B that make an angle ϕ when placed tail to tail is A× B º AB cos f. © 2015 Pearson Education, Inc. Slide 10-79 Chapter 10: Summary Concepts: Projectile motion in two dimensions • For a projectile that moves near Earth’s surface and experiences only the force of gravity, the acceleration has a magnitude g and is directed downward. • The projectile’s horizontal acceleration is zero, and the horizontal component of its velocity remains constant. • At the highest point in the trajectory of a projectile, the vertical velocity component υy is zero, but the vertical component of the acceleration is g and is directed downward. © 2015 Pearson Education, Inc. Slide 10-80 Chapter 10: Summary Quantitative Tools: Projectile motion in two dimensions • If a projectile is at a position (x, y), its position vector r is r = xiˆ + yĵ. • If a projectile has position components xi and yi and velocity components υx,i and υy,i at one instant, then its acceleration, velocity, and position components a time interval Δt later are ax = 0 ay = – g u x ,f = u x ,i u y ,f = u y ,i – gDt xf = xi + u x ,i Dt yf = yi + u y ,i Dt – 12 g(Dt)2 © 2015 Pearson Education, Inc. Slide 10-81 Chapter 10: Summary Concepts: Collisions and momentum in two dimensions • Momentum is a vector, so in two dimensions its changes in momentum must be accounted for by components. • This means two equations, one for the x component and one for the y component of momentum. • The coefficient of restitution is a scalar and is accounted for by a single equation. © 2015 Pearson Education, Inc. Slide 10-82 Chapter 10: Summary Quantitative Tools: Collisions and momentum in two dimensions Δpx =Δp1x + Δp2x = m1(υ1x,f – υ1x,i) + m2(υ2x,f – υ2x,i) = 0 Δpy =Δp1y + Δp2y = m1(υ1y,f – υ1y,i) + m2(υ2y,f – υ2y,i) = 0 © 2015 Pearson Education, Inc. Slide 10-83 Chapter 10: Summary Concepts: Forces in two dimensions • In two-dimensional motion, the component of the acceleration parallel to the instantaneous velocity changes the speed and the component of the acceleration perpendicular to the instantaneous velocity changes the direction of the instantaneous velocity. • When choosing a coordinate system for a problem dealing with an accelerating object, if possible make one of the axes lie along the direction of the acceleration. © 2015 Pearson Education, Inc. Slide 10-84 Chapter 10: Summary Concepts: Friction • When two surfaces touch each other, the component of the contact force perpendicular (normal) to the surfaces is called the normal force and the component parallel (tangential) to the surfaces is called the force of friction. • The direction of the force of friction is such that the force opposes relative motion between the two surfaces. When the surfaces are not moving relative to each other, we have static friction. © 2015 Pearson Education, Inc. Slide 10-85 Chapter 10: Summary Concepts: Friction • When the forces are moving relative to each other, we have kinetic friction. • The magnitude of the force of kinetic friction is independent of the contact area and independent of the relative speeds of the two surfaces. © 2015 Pearson Education, Inc. Slide 10-86 Chapter 10: Summary Quantitative Tools: Friction • The maximum magnitude of the force of static friction between any two surfaces 1 and 2 is proportional to the normal face: (F12s )max = ms F12n , where μs is the unitless coefficient of static friction. This upper limit means that the magnitude of the frictional force must obey the condition F12s £ ms F12n . • The magnitude of the force of kinetic friction is also proportional to the normal face: F12k £ mk F12n , where μk ≤ μs is the unitless coefficient of kinetic friction. © 2015 Pearson Education, Inc. Slide 10-87 Chapter 10: Summary Concepts: Work • For a sliding object, the normal force does no work because this force is perpendicular to the direction of the object’s displacement. • The force of kinetic friction is a nonelastic force and thus causes energy dissipation. • The force of static friction is an elastic force and so causes no energy dissipation. • The work done by gravity is independent of path. © 2015 Pearson Education, Inc. Slide 10-88 Chapter 10: Summary Quantitative Tools: Work • The work done by a constant nondissipative force when the point of application of the force undergoes a displacement DrF is W = F × DrF . • The work done by a variable nondissipative force when the point of application of the force undergoes a displacement is W = ò rrf F(r ) × dr . i • This is the line integral of the force over the path traced out by the point of application of the force. © 2015 Pearson Education, Inc. Slide 10-89 Chapter 10: Summary Quantitative Tools: Work • For a variable dissipative force, the change in the thermal energy is r DEth = – ò rf F(rcm ) × drcm . i • When an object descends a vertical distance h, no matter what path it follows, the work gravity does on it is W = mgh. © 2015 Pearson Education, Inc. Slide 10-90